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x + y = x, solve for X (pg. 6)
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| astroboy |
| quote: | Originally posted by Nadi
x is undefined and any non imaginary number. |
Why does it have to be non imaginary?
if x = i [where i = (-1)^1/2]
then
i + y = i
y = i - i
y = 0
Still seems to work...
PS - take it easy on me, I haven't touched complex numbers since high school. |
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| DrUg_Tit0 |
| quote: | Originally posted by Galapidate
f(x) is the same thing as 'y'. Thus, f(x)+x=x is the same as y+x=x, as we've been saying so far :P |
Well, duh, I know that. I just mentioned that there was something about such a problem mentioned recently that was not very trivial. I remembered what it was now, and it was about reiterating a function to approximate its value. This was mentioned as a specific case in which there was some additional step taken to solve it. Like, if you have the function y=f(x), you'd replace y with x1, and you'd get x1=f(x). Then you'd continue and get x2=f(x1) and so on. Each step would bring you closer to the solution But if the function was of the type f(x)-x, then you'd get f(x)=0. Not much to do with this specific problem though, as this one is elementary. |
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| Lover Boy |
| quote: | Originally posted by whiskers
i own you all, i will derive and antidifferentiate songs and i will sell them on napster to stupid people and i'll become rich MUHAHAHAHAHA |
Antidifferentiate eh? I'm quite sure that's called Integration m8 ;) & if u differentiated then integrated a fucnction you'd arrive back at the original function. So ull b sellin the exact same song 2 ppl. Or wos that ur point? |
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| drizzt81 |
| quote: | Originally posted by placebo
x + y = x
solve for x
anyone?? |
your equation is solved for x already.
x = x + y, therefore y = 0.
gees... some people. |
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| mezzir |
thread of the week!
btw we need more simple problems to overanalyze
ooh wait nm, i got it
.99999(repeating) = 1
discuss. |
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| DrUg_Tit0 |
| quote: | Originally posted by mezzir
thread of the week!
btw we need more simple problems to overanalyze
ooh wait nm, i got it
.99999(repeating) = 1
discuss. |
Let's take the following sequence 0.9, 0.99, 0.999... Now, this sequence is converging and equal to 1 since for every e>0 exists a n0(e) from N so that for every n>n0 |an-1| |
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| reveal |
| quote: | Originally posted by dj_mdma
The square root of a minus number doesn't work on a calculator, but it still exists.
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depends on the calculator, mine handles imaginary numbers ;)
and for the record, i^2 = -1, thus sqrt(-x^2) = x*i
(sorry if i'm repeating things, haven't read the whole thread) |
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| mezzir |
| quote: | Originally posted by DrUg_Tit0
Let's take the following sequence 0.9, 0.99, 0.999... Now, this sequence is converging and equal to 1 since for every e>0 exists a n0(e) from N so that for every n>n0 |an-1| |
huh, never seen it done that way
one easy way - between any two real numbers, there must be another real number, therefore .999...must = 1
my way that i came up with right as a friend posed the question to me a few years back - .111111... = 1/9, right? and .22222... = 2/9, right? well then wouldn't .999999... = 9/9? and 9/9 = 1
voila |
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| Nadi |
| quote: | Originally posted by astroboy
Why does it have to be non imaginary?
if x = i [where i = (-1)^1/2]
then
i + y = i
y = i - i
y = 0
Still seems to work...
PS - take it easy on me, I haven't touched complex numbers since high school. |
Hmm Your right :P
and drizzt your solving for y not x :eek: |
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| DrUg_Tit0 |
| quote: | Originally posted by mezzir
huh, never seen it done that way
one easy way - between any two real numbers, there must be another real number, therefore .999...must = 1
my way that i came up with right as a friend posed the question to me a few years back - .111111... = 1/9, right? and .22222... = 2/9, right? well then wouldn't .999999... = 9/9? and 9/9 = 1
voila |
You wanted the overanalyze the problem, didn't you? I assure you my solution is correct, yet unnecessarily complicated. |
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| placebo |
Holy crap, I come back and check the chillout room, and this thread has reached 5 pages....
But does anyone have the answer yet? |
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| DrUg_Tit0 |
| quote: | Originally posted by placebo
But does anyone have the answer yet? |
Yes, it's been solved for like 25 times already. y is not a function of x, and therefore x+y=x is true for every x e Z. Meaning all the numbers you can think of for x are equally valid solutions of the equation. y however, is a fixed nomber, and it is equal to zero. |
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